re: Solve this OT

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I do not have a choice of what box to pick the second ball from

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correct, but prior to your second pick, you don't know which box you have. Technically the options of both boxes are still available.

quote:

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That option was never ruled out

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Ok now you are just trolling.

It isn’t possible to be this stupid.


Maybe take a break and read the original question again....








Slowly

5 more pages

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RobbBobb

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Clown.

I'm not reading 10 pages, but I'm going on record and saying the OT is fricking stupid and I'm guessing that the people who make the OT fricking stupid were shown why they're wrong on page 3, but have been arguing why they are right since then.

ETA: How close am I?

The mother was the doctor...

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There's a stronger likelihood that you chose the box with two golds in the beginning. That gets factored in.

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Once you pick the gold ball, youve eliminated the AS box from all future probabilities. So your next pick establishes a new probability. You are down to 2 options, not the possibilities you started with

If the opposing coach tells you that in this series of downs we are going to run right (AS), run left (AG), and run up the middle (M). And on 1st down he ran the ball to the right (AS), then your next defensive call adjusts to the only 2 options left. Because option 1 is no longer a factor in your decision making

As always, it's time to get all Monte Carlo up in this bitch.
This is in Python, and the leading dots on some of the lines just represent spaces that won't format correctly here any other way I've found.


import random
# Three boxes in the 'box' list, each box is
# a list of two balls.

box = [['G','G'],['G','S'],['S','S']]

def choose_ball ():
# We'll choose a box at random (and remember it)
# and choose a ball from it.
. this_box = random.randrange (3)
. return (this_box, random.choice (box [this_box]))

def other_ball (box_idx):
# If we bother to look at the 'other' ball, we know
# we have already chosen a 'G', so we must be in
# either box [0] or box [1]
# in box [0], the other ball has to be 'G',
# in box [1], the other ball has to be 'S'

. if box_idx == 0:
.. return 'G'
. if box_idx == 1:
.. return 'S'

trials = 0
golds = 0

# Now we do the test a million times
for i in range (1000000):
# Pick a ball and remember which box we're in
. my_box, my_ball = choose_ball ()
. if my_ball == 'S':
# Silver ball first -- don't count anything
# Just do the next one
.. continue
. else:
# Gold ball first, so we count this one
# and look at the other ball in the box.
.. trials = trials + 1
.. if other_ball (my_box) == 'G':
... golds = golds + 1

print (float (golds) / float (trials))


So, for each test, we pick a ball 1 million times. If we pick a silver ball, we go on to the next pick. If we pick a gold ball, we look at the other ball in the box we picked from. At the end we look at ONLY the times we picked a gold ball first, and we see the fraction of times that the other ball was also gold. It's fairly fast on a modern PC, so I ran it 10 times (remember it was a million picks for each run and another pick if we picked gold first).

Results are:

0.665746226428
0.667399396348
0.667402563641
0.667849064158
0.666473975721
0.665929128366
0.666613307294
0.666898288197
0.667379593472
0.666684691288

It's pretty safe to say that simulation agrees with the assertion that your chances are 2/3 to pick a second gold ball GIVEN that you've picked a gold ball first.

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If the opposing coach tells you that in this series of downs we are going to run right (AS), run left (AG), and run up the middle (M). And on 1st down he ran the ball to the right (AS), then your next defensive call adjusts to the only 2 options left. Because option 1 is no longer a factor in your decision making

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Uh, Roll Tide?

Monty Hall makes perfect sense but still blows my mind.